Balance the equation first:
2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)
Moles of KClO3 = 110 / 122.5 = 0.89
Following the balanced chemical equation:
We can say moles of O2 produce =

x moles of KClO3
So, O2 = (3 / 2) x 0.89
= 1.34 moles
So, Volume at STP = nRT / P
T = <span>273.15 K
P = 1 atm
So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>
Answer:
4L
Explanation:
To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.
From the question given above,
Mass of O2 = 5.72g
Molar Mass of O2 = 32g/mol
Number of mole =Mass/Molar Mass
Number of mole of O2 = 5.72/32
Number of mole of O2 = 0.179 mole
Now, we can calculate the volume of O2 at stp as follow:
1 mole of a gas occupy 22.4L at stp.
Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L
Therefore, the volume occupied by the sample of O2 is 4L
2C2H5OH + 2Na ---- 2C2H5ONa + H2